Question

i inserted a picture of the question 6, i can give you the answer to the previous question if it helps

Answer 1

For

$w^2-3w=350$

To be a perfect square trinomial, we need to find a such that

$\begin{gathered} (w-a)^2=w^2-3w+b \\ b\to\text{constant} \end{gathered}$

Thus,

$(w-a)^2=w^2-2aw+a^2$

Then,

$\begin{gathered} \Rightarrow w^2-2aw+a^2=w^2-3w+b \\ \Rightarrow-2a=-3 \\ \Rightarrow a=\frac{3}{2} \end{gathered}$

Furthermore,

$\begin{gathered} \Rightarrow a^2=b \\ \Rightarrow\frac{9}{4}=b \\ \end{gathered}$

We need to add 9/4 to both sides of the equation, resulting in

$\begin{gathered} w^2-3w+\frac{9}{4}=350+\frac{9}{4} \\ \Rightarrow(w-\frac{3}{2})^2=\frac{1409}{4} \end{gathered}$